# Relating integral of powers of Sin b/w 0 and pi/2 to factorial form

1. Aug 13, 2014

### musik132

Our integral
$$\int\limits_0^{\pi/2} \sin^{2a+1}(x)\,dx$$

Has a Factorial Form:
$${(2^a a!)}^2 \over (2a+1)!$$

What is the process behind going from that integral to that factorial form?

My approach which is not very insightful:
I used mathematica to calculate the integral to return:
$$\pmb{\frac{\sqrt{\pi } \text{Gamma}[1+a]}{2 \text{Gamma}\left[\frac{3}{2}+a\right]}}$$
I know Gamma[1+a] = a! and Gamma[3/2+a] has a factorial form also but doesn't help me to reduce to that form.

Griffiths just says that integral equals (2*4*...2a)/(1*3*5...*[2a+1]) to get from this to that factorial form is easy but I got lost in his integration.

Last edited: Aug 13, 2014
2. Aug 13, 2014

### micromass

Staff Emeritus
What do you get after integrating by parts twice?

3. Aug 13, 2014

### musik132

$$\pi/2-(2a+1)\int\limits_0^{\pi/2} \sin^{2a}(x)cos(x)x$$

Sorry I still don't see how to finish the connection.

Edit: Didn't see you said twice IBP ill go back and retry this

4. Aug 13, 2014

### musik132

taking u = sin^2a(x) and v'=xcos(x) , I get:
$$-a\pi+(2a+1)(2a)\int\limits_0^{\pi/2} Sin^{2a}(x)x +Sin^{2a-1}(x)cos(x)dx$$

Sadly my math isn't great and cant seem to figure out how this would lead to the factorial form.

So I tried to integrate by parts again and try to simplify and it just got really messy.
I tried to take u = sin^2a(x)cos(x) and v'=x and that got messy also compared to the one above so I didn't pursue it.

Last edited: Aug 13, 2014